 Solution

For $n=1.5$, the critical angle is $i_c=\arcsin (1/1.5)\simeq 41.8103^{\circ }$. Here $0\le r<i_c<\alpha =60^{\circ }$, so the condition at B is $\alpha -r>i_c$. At the threshold,Thus the requested positive-angle interval for strict total internal reflection isAt the upper equality the transmitted ray is grazing; normal incidence $\theta =0$ also gives total internal reflection at B if the endpoint is included. The distinction follows the question's positive-angle convention.